Question

If `f(x) = lnx + 2`, what is the tangent to the graph at `x = 1`?

Answer 1

The tangent has equation `y = x+ 1`.

Explanation:

The point of tangency will be

`f(1) = 2 + ln(1) = 2 + 0 = 2`

The derivative will give us the slope of the tangent at any point in its domain.

`f'(x) = 1/x`

So the slope at `x= 1` is `1/1 = 1`.

`y - y_1 = m(x - x_1)`

`y - 2 = 1(x - 1)`

`y - 2 = x - 1`

`y = x + 1`

We can do a graphical verification to confirm.

Hopefully this helps!

Answer 2

Calculate the derivative at `x=1`, find `f(1)`, and use these to construct a point slope equation with the derivative as the slope. Convert to slope intercept if desired. `y-2= x-1 -> y = x+1`

Explanation:

We are given `f(x)=2+lnx, x_0=1`. Then `f(1) = 2 + ln 1 = 2`, meaning that the point (1, 2) is on the curve. We will, need this later.

In order to find our tangent line equation, we must first find the derivative of f(x), which will be the formula for the slope of the line tangent to the curve at any point on the curve. We will then find the derivative at x=1, to find the slope of the line tangent to the graph of the curve at x=1. Finally, we will use the point we found earlier and the slope to put the tangent line into point-slope form and then slope-intercept form.

The derivative of f(x) is `d/dxf(x) = d/dx(2+lnx) = 1/x = f'(x)`

Plugging in `x_0`, we find `f'(x_0) = f'(1)= 1/1 =1`

With this and the point we found, we find the point slope equation:

`y-y_0=f'(1)(x-x_0) -> y-2 = 1(x-1) - > y-2 = x-1`

And if we want slope intercept form...

`-> y = x+1`